数理解析研究所講究録別冊 = RIMS Kokyuroku Bessa (2011), B26:

Σχετικά έγγραφα
Second Order Partial Differential Equations

2 Composition. Invertible Mappings

Uniform Convergence of Fourier Series Michael Taylor

Section 8.3 Trigonometric Equations

Example Sheet 3 Solutions

Solutions to Exercise Sheet 5

Ordinal Arithmetic: Addition, Multiplication, Exponentiation and Limit

EE512: Error Control Coding

C.S. 430 Assignment 6, Sample Solutions

ST5224: Advanced Statistical Theory II

Congruence Classes of Invertible Matrices of Order 3 over F 2

Reminders: linear functions

Jesse Maassen and Mark Lundstrom Purdue University November 25, 2013

CHAPTER 25 SOLVING EQUATIONS BY ITERATIVE METHODS

Every set of first-order formulas is equivalent to an independent set

Lecture 2: Dirac notation and a review of linear algebra Read Sakurai chapter 1, Baym chatper 3

REMARK ON THE OPTIMAL REGULARITY FOR EQUATIONS OF WAVE MAPS TYPE

Coefficient Inequalities for a New Subclass of K-uniformly Convex Functions

HOMEWORK 4 = G. In order to plot the stress versus the stretch we define a normalized stretch:

b. Use the parametrization from (a) to compute the area of S a as S a ds. Be sure to substitute for ds!

1 String with massive end-points

Homework 8 Model Solution Section

The Pohozaev identity for the fractional Laplacian

Areas and Lengths in Polar Coordinates

Spherical Coordinates

Matrices and Determinants

Parametrized Surfaces

Areas and Lengths in Polar Coordinates

Section 7.6 Double and Half Angle Formulas

Math221: HW# 1 solutions

Practice Exam 2. Conceptual Questions. 1. State a Basic identity and then verify it. (a) Identity: Solution: One identity is csc(θ) = 1

Phys460.nb Solution for the t-dependent Schrodinger s equation How did we find the solution? (not required)

Nowhere-zero flows Let be a digraph, Abelian group. A Γ-circulation in is a mapping : such that, where, and : tail in X, head in

D Alembert s Solution to the Wave Equation

A Note on Intuitionistic Fuzzy. Equivalence Relation

Απόκριση σε Μοναδιαία Ωστική Δύναμη (Unit Impulse) Απόκριση σε Δυνάμεις Αυθαίρετα Μεταβαλλόμενες με το Χρόνο. Απόστολος Σ.

Chapter 6: Systems of Linear Differential. be continuous functions on the interval

4.6 Autoregressive Moving Average Model ARMA(1,1)

derivation of the Laplacian from rectangular to spherical coordinates

Statistical Inference I Locally most powerful tests

Boundedness of Some Pseudodifferential Operators on Bessel-Sobolev Space 1

Fourier Series. MATH 211, Calculus II. J. Robert Buchanan. Spring Department of Mathematics

SCHOOL OF MATHEMATICAL SCIENCES G11LMA Linear Mathematics Examination Solutions

Space-Time Symmetries

3.4 SUM AND DIFFERENCE FORMULAS. NOTE: cos(α+β) cos α + cos β cos(α-β) cos α -cos β

6.3 Forecasting ARMA processes

Iterated trilinear fourier integrals with arbitrary symbols

SCITECH Volume 13, Issue 2 RESEARCH ORGANISATION Published online: March 29, 2018

Partial Differential Equations in Biology The boundary element method. March 26, 2013

New bounds for spherical two-distance sets and equiangular lines

A General Note on δ-quasi Monotone and Increasing Sequence

Exercises to Statistics of Material Fatigue No. 5

ANSWERSHEET (TOPIC = DIFFERENTIAL CALCULUS) COLLECTION #2. h 0 h h 0 h h 0 ( ) g k = g 0 + g 1 + g g 2009 =?

w o = R 1 p. (1) R = p =. = 1

CRASH COURSE IN PRECALCULUS

The Negative Neumann Eigenvalues of Second Order Differential Equation with Two Turning Points

Integrals in cylindrical, spherical coordinates (Sect. 15.7)

Exercises 10. Find a fundamental matrix of the given system of equations. Also find the fundamental matrix Φ(t) satisfying Φ(0) = I. 1.

Homework 3 Solutions

forms This gives Remark 1. How to remember the above formulas: Substituting these into the equation we obtain with

Srednicki Chapter 55

Problem Set 9 Solutions. θ + 1. θ 2 + cotθ ( ) sinθ e iφ is an eigenfunction of the ˆ L 2 operator. / θ 2. φ 2. sin 2 θ φ 2. ( ) = e iφ. = e iφ cosθ.

Problem Set 3: Solutions

The Simply Typed Lambda Calculus

( y) Partial Differential Equations

Fractional Colorings and Zykov Products of graphs

Homomorphism in Intuitionistic Fuzzy Automata

Second Order RLC Filters

Other Test Constructions: Likelihood Ratio & Bayes Tests

Tridiagonal matrices. Gérard MEURANT. October, 2008

SOME PROPERTIES OF FUZZY REAL NUMBERS

DESIGN OF MACHINERY SOLUTION MANUAL h in h 4 0.

Econ 2110: Fall 2008 Suggested Solutions to Problem Set 8 questions or comments to Dan Fetter 1

Chapter 6: Systems of Linear Differential. be continuous functions on the interval

A Two-Sided Laplace Inversion Algorithm with Computable Error Bounds and Its Applications in Financial Engineering

Lecture 15 - Root System Axiomatics

n=2 In the present paper, we introduce and investigate the following two more generalized

LOW REGULARITY SOLUTIONS OF THE CHERN-SIMONS-HIGGS EQUATIONS IN THE LORENTZ GAUGE

Numerical Analysis FMN011

On a four-dimensional hyperbolic manifold with finite volume

Approximation of distance between locations on earth given by latitude and longitude

ECE Spring Prof. David R. Jackson ECE Dept. Notes 2

9.09. # 1. Area inside the oval limaçon r = cos θ. To graph, start with θ = 0 so r = 6. Compute dr

MINIMAL CLOSED SETS AND MAXIMAL CLOSED SETS

Notes on the Open Economy

Lecture 26: Circular domains

6.1. Dirac Equation. Hamiltonian. Dirac Eq.

Strain gauge and rosettes

Inverse trigonometric functions & General Solution of Trigonometric Equations

Homomorphism of Intuitionistic Fuzzy Groups

SOLUTIONS TO MATH38181 EXTREME VALUES AND FINANCIAL RISK EXAM

PROPERTIES OF CERTAIN INTEGRAL OPERATORS. a n z n (1.1)

Lecture 2. Soundness and completeness of propositional logic

12. Radon-Nikodym Theorem

de Rham Theorem May 10, 2016

Finite Field Problems: Solutions

Solution Series 9. i=1 x i and i=1 x i.

Lecture 21: Properties and robustness of LSE

= {{D α, D α }, D α }. = [D α, 4iσ µ α α D α µ ] = 4iσ µ α α [Dα, D α ] µ.

Answer sheet: Third Midterm for Math 2339

Transcript:

Title On a Bilinear Estimate of Schroding Analysis and Nonlinear Partial Diff Author(s) Fang, Yung-fu Citation 数理解析研究所講究録別冊 = RIMS Kokyuroku Bessa (011), B6: 1-14 Issue Date 011-05 URL http://hdl.handle.net/433/187887 Right Type Departmental Bulletin Paper Textversion publisher Kyoto University

On a Bilinear Estimate of Schrödinger Waves Yung-fu Fang Abstract In this paper we want to consider a bilinear space-time estimate for homogeneous Schrödinger equations. We give an elementary proof for the estimates in Bourgain space, which is in a form of scaling invariance. 1 Introduction Consider the homogeneous Schrödinger equations { iut u = 0, (x, t) R n R, u(0) = f; { ivt v = 0, (x, t) R n R, v(0) = g. (1.1) Via Fourier transform, the solution u and v can be written as u(t) = e it f and v(t) = e it g. (1.) Thus to study the estimates of the product of Schrödinger waves, uv, is to study the estimates of the product e it fe it g. (1.3) There are many literature investigating on the topic of bilinear estimates for Schödinger waves. In 98, Ozawa and Tsutsumi [OT] proved an L estimate for u v with 1 derivative for n = 1, ( ) 1 4 (e it f)(e it g) = 1 f L L g L. (1.4) t,x In 98, Bourgain [Bo] showed a refinements of Strichartz inequality for n =. supported on ξ N, ĝ is supported on ξ M, and M << N, then (e it f)(e ±it g) L t,x If f is ( M N ) 1 f L g L. (1.5) In 01, Kenig etc. [CDKS] obtained a bilinear estimate in Bourgain space for nonlinear Schrödinger equation in two dimension. Let b = 1+. If 1 (1 b) < s and σ < min ( s + 4 1, s + (1 b)), then uv X σ,b1 u X s,b v X s,b. (1.6) In 03, Tao [T1] obtained a sharp bilinear restriction estimate for paraboloids. Let q > n + 3 n + 1, n, N > 0, and f and g have Fourier transform supported in the region ξ N. Suppose 1

that dist(supp f, supp ĝ) cn. Then we have e it fe it g n+ n L q N q f t,x (Rn+1 ) L g L. (1.7) In 05, Burq, Jérard, and Tzvetkov [BGT] derived bilinear eigenfunction estimates on sphere and on Zoll surfaces. In 09, Keraani and Vargas [KV] showed a bilinear estimate of uv in L n+ n norm, where n. If b (0, ), then n+ e it fe it g L n+ C f n (R n+1 ) Ḣ b g Ḣ1b. (1.8) In 09, Kishimoto [K] derived an improved bilinear estimate for quadratic Schrödinger equation in one and two dimensions. The estimate is in a variant of Bourgain space with weighted norm. In 10, Chae, Cho, and Lee [CCL] proved an interactive estimate of uv in a mixed norm. Let n. If = n(1 1 ), 1 < r, q > 1, s < 1 1, then q r r e it fe it g L q C f Ḣs g Ḣs. (1.9) t Lr x Let D, S +, and S be the operators with the symbols respectively. D = ξ, Ŝ + = τ + ξ, and Ŝ = τ ξ, (1.10) Theorem 1. Let n. If for j = 1,, β 0 + β + + β + n = α 1 + α, β 0, β α j + n 1 0, (β, α j ) (0, n 1 ), and β 0 > n 1, (1.11) then ) D β 0 S β + + S β (e it fe it g C f Ḣα L(Rn+1) 1 (R n ) g Ḣ α (R n ). (1.1) Notice that Strichartz Estimate for Homogeneous Schrödinger equation for n = reads which coincides with the bilinear estimate when u L 4 f L, u L 4 = uu L f L f L, β 0 = β + = β = α 1 = α = 0. The estimate is given in the form of scaling invariance. The conditions stated in the theorem come from the scaling invariance and interactions between frequencies.

The proof of Theorem 1 is based on the ideas of the work of Foschi and Klainerman [FK], and the work of Klainerman and Machedon [KM], however some modifications for adapting the case of Schrödinger are required. The purpose of this work is to derive a new estimate with an elementary proof. The paper is organized as follows: In Section, we prove Theorem 1. In Section 3, we state and prove some properties which are the technical parts left in the proof of Theorem 1. Bilinear Estimates for Schrödinger waves We denote the Fourier transform of the function u(t, x) by û(t, ξ) with respect to the space variable and by ũ(τ, ξ) with respect to the space-time variables. For simplicity, we call We now prove Theorem 1. D β 0 Ŝ β + + Ŝ β. Proof. First we compute the Fourier transform of the product e it fe it g with respect to space variables, ê it f êit g(ξ) = e it ξη f(ξ η)e it η ĝ(η)dη = e it( ξη + η ) f(ξ η)ĝ(η)dη. Thus its Fourier transform with respect to space-time variables is δ(τ ξ η η ) f(ξ η)ĝ(η)dη, where δ(τ ξ η η )dη is viewed as a measure supported on surfaces {η : τ = ξ η + η }. We split the integral into three parts in the following way. First we ine a function h(γ) = γ 1 γ (.1) which will appear in the proof later, see figure 1. Since the equation h(γ) = 1/3 has two roots 9 ± 6, we denote the two roots by γ 1 = 9 6 and γ = 9 + 6. Then we decompose the η-space into S c, see figure, where = {η : 1 ξ ξ η + η γ 1 ξ }, = {η : γ 1 ξ ξ η + η γ ξ }, and S c = {η : γ ξ ξ η + η }. (.) Thus we have 3

Figure 1: = = D β 0 S β + + S β + Figure : (e it fe g) it L δ(τ ξ η η ) f(ξ η)ĝ(η)dη dτdξ + δ(τ ξ η η ) f(ξ η)ĝ(η)dη S c dτdξ. (.3) Hence it is sufficient to bound each of the above integrals. For simplicity, we denote Φ(η) τ ξ η η. Using Hölder inequality, we can bound the first integral in (.3) as follows. δ(τ ξ η η ) f(ξ η)ĝ(η)dη dτdξ δ(φ(η)) ξ η α 1 η α dη δ(φ(η)) ξ η α 1 f(ξ η) η αĝ(η) dηdτdξ { ξ δ(φ(η))dτ} η α 1 f(ξ η) η αĝ(η) dηdξ C f Ḣα1 g Ḣα, provided that δ(φ(η)) ξ η α 1 η α dη C, for all τ, ξ. (.4) 4

For the second integral we can get the desired bound in the same vain, δ(φ(η)) f(ξ η)ĝ(η)dη dτdξ C f Ḣα1 g Ḣα, (.5) provided that δ(φ(η)) ξ η α 1 η α dη C, for all τ, ξ. (.6) Notice that we have ξ η η and ξ ϕ ϕ on the set S c. Using the fact that z = z z, the Fubini theorem, and change of variables, we can bound the third integral in (.3) as follows. δ(φ(η)) f(ξ η)ĝ(η)dη dτdξ S c = δ(φ(η)) f(ξ η)ĝ(η)dη δ(φ(ϕ)) f(ξ ϕ)ĝ(ϕ)dϕdτdξ = δ(φ(η) Φ(ξ ϕ)) f(ξ η)ĝ(η) f(ϕ)ĝ(ξ ϕ)dϕdηdξ ( δ(φ(η) Φ(ξ ϕ)) 1/ ϕ α 1 ( ϕ η ) α 1+α f(ϕ) η αĝ(η) dϕdηdξ) ( δ(φ(η) Φ(ξ ϕ)) 1/ ξ ( ξ ϕ ξ η ) α 1+α ϕ α 1 f(ξ ϕ) ξ η αĝ(ξ η) dϕdηdξ). Hence we have the following bound δ(φ(η) Φ(ξ ϕ)) ( ϕ η ) α 1+α dξ ϕ α 1 f(ϕ) η αĝ(η) dϕdη C f Ḣ α 1 g, Ḣ α T c provided that T c δ(φ(η) Φ(ξ ϕ)) ( ϕ η ) α 1+α dξ C, for all ϕ and η, (.7) where T c = {ξ : ξ ϕ + ϕ, ξ η + η γ ξ } and τ = ξ ϕ + ϕ = ξ η + η. Therefore the proof of the Theorem is complete once the claims, (.4), (.6), and (.7) are proved. Remark 1. What left to be done are the following estimates. Claims: There is a constant C which is independent of τ, ξ, ϕ, and η such that the following inequalities hold. D β 0 Ŝ β δ(τ ξ η η ) + + Ŝβ ξ η α 1 η α dη C, for all τ, ξ. (.8) D β 0 Ŝ β δ(τ ξ η η ) + + Ŝβ ξ η α 1 η α dη C, for all τ, ξ. (.9) 5

Dβ 0 Ŝ β + + Ŝβ T c δ( ξ ϕ + ϕ ξ η η ) ( ϕ η ) α 1+α dξ C, for all ϕ and η, (.10) where τ = ξ ϕ + ϕ = ξ η + η. The proofs of the above claims will be given in the next section. 3 Proofs of Claims Now we are ready to prove the claims. First we prove the claims which come from the proof of bilinear estimates for uv. 1 ξ η + ξ η γ 1 ξ }. If β 0 + β + + β + Lemma 1 (Claim (.8)). Let = {η : n = α 1 + α and n, then D β 0 Ŝ β + + Ŝβ for all τ and ξ. ( Proof. We set ζ = R the indentities Then we use spherical coordinates δ(τ ξ η η ) ξ η α 1 η α dη C, (3.1) ), where R is the rotation such that Rξ = ξ e 1, then we have η ξ ξ η = ζ 1 ξ e 1 and η = ζ + 1 ξ e 1. (3.) ζ = (X 1,, X n ) = ρ(cos φ, sin φω ) = ρω, (3.3) where ω S n and ω S n1, so that we can rewrite the integral in (3.1) as δ(τ ξ η η ) 1 ρ0 n1 ξ η α 1 η α dη = ξ η α 1 η α dω, (3.4) 4ρ 0 where ρ 0 = 1 4 (τ ξ ). Using the identity dω = (sin φ) n dφdω, the identities for ξ η and η in (3.), and the change of variables p = cos φ, we can simplify the integral further. 1 1 ξ η α 1 η α ρ n 0 sin n φdφdω ρn 0 (1 p ) (n3)/ τ α 1+α (1 + λp) α 1 (1 λp) α dp, (3.5) 1 where λ = ξ ρ 0 = ξ τ ξ. Notice that 0 λ 1 τ τ 3 ξ η + η for η. under the restriction τ = 6

We set τ γ 1 = γ ξ which implies that λ = = h(γ). Then we have 1 γ γ γ 1 which implies that τ ξ, and ρ 0 = 1 γ 1 ξ ξ. Thus we can estimate the quantity in (3.1) as follows. D β 0 Ŝ β δ(τ ξ η η ) + + Ŝβ ξ η α 1 η α dη ( ) n ξ β 0+4β + +4β α 1 α γ β γ 1 ξ 1 (1 p ) (n3)/ 1 (1 + λp) α 1 (1 λp) α dp. The above quantity is bounded if we require that n and β 0 + β + + β + n = α 1 + α. 1 Lemma (Claim (.9)). Let = {η : γ 1 ξ η + ξ η γ ξ }. If n, β α j n 1 for j = 1,, then D β 0 Ŝ β + + Ŝβ β 0 + β + + β + n for all τ and ξ. ( Proof. As in the proof of Lemma 1, we set ζ = R and = α 1 + α,, β 0, and (β, α j ) (0, n 1 ), δ(τ ξ η η ) ξ η α 1 η α dη C, (3.6) η ξ ), where the rotation Rξ = ξ e 1, ζ = (X 1,, X n ) = ρ(cos φ, sin φω ) = ρω. (3.7) Using the above, the identity dω = (sin φ) n dφdω, and the identities for ξ η and η, we can rewrite the integral in (3.6) as δ(τ ξ η η ) ξ η α 1 η α dη ρn 0 τ α 1+α 1 1 (1 p ) (n3)/ (1 + λp) α 1 (1 λp) α dp, (3.8) where ρ 0 = 1 4 (τ ξ ), the change of variables p = cos φ, and λ = ξ ρ 0 τ Again we set τ = γ ξ and then we have γ 1 γ γ and 1 3 τ = ξ η + η for η. These imply that = ξ τ ξ. τ λ 1 under the restriction τ ξ, (γ 1) (1 λ), and ρ 0 = 1 γ 1 ξ ξ. (3.9) 7

Now we can combine the above observations to simplify the quantity in (3.1) as follows. D β 0 Ŝ β δ(τ ξ η η ) + + Ŝβ ξ η α 1 η α dη ξ β 0+4β + +4β +nα 1 α γ 1 β 1 1 (1 p ) (n3)/ (1 + λp) α 1 (1 λp) α dp. To bound the above integral we split it into two parts, one is over [-1, 0] while the other is over [0, 1]. Since the estimates for the two parts are the same, thus we only prove the second part. First we have 1 0 (1 p ) (n3)/ 1 (1 + λp) α 1 (1 λp) α dp 0 (1 p) (n3)/ dp (3.10) (1 λp) α Using the change of variables p = (1 λ)q + λ, the integral is changed into (1 λ) n3 +1α λ 1λ 1 (1 + q) (n3)/ (1 + λ + λq) α dq. Again we split the above integral into two parts. For the first part, we have 1 λ 1λ 1 1 (1 + q) (n3)/ (1 + λ + λq) α dq (1 + q) (n3)/ dq C, (1 + λ + λq) α provided that n. For the second part, we have (1 λ) n1 +α for α < n1, log(1 λ) for α = n1, C for α > n1, Thus we get 1 (1 λ) β (1 p ) (n3)/ 0 (1 + λp) α 1 (1 λp) α dp (1 λ) β + n1 α (1 λ) n1 +α for α < n1, C(1 λ) β + n1 α + (1 λ) β + n1 α log(1 λ) for α = n1, (1 λ) β + n1 α C for α > n1, Hence the above integral is bounded if (β, α ) is in the set S 1 (S S 3 S 4 ), where S 3 S 1 = {(β, α ) : β α n 1 }, S = {(β, α ) : β 0, α < n 1 }, ={(β, α ): β > α n 1, α = n 1 }, and S 4 8 = {(β, α ): β α n 1, α > n 1 }.

Therefore the quantity in (3.6) is bounded if we require that β 0 + β + + β + n = α 1 + α, n, β α n 1, β 0, and (β, α ) (0, n 1 ). The conditions for α 1 is the same as that of α. This completes the proof. Remark. Notice that for the integral over S c we have 0 λ 1, γ 3 γ, τ = γ ξ, and ρ 0 = γ 1 ξ / γ ξ. If we follow the same path to estimates it, then we obtain D β 0 Ŝ β δ(τ ξ η η ) + + Ŝβ ξ η α 1 η α dη ξ β 0+4β + +4β +nα 1 α γ β ++β + n α 1α 1 1 (1 u ) n3 (1 + λp) α 1 (1 λp) α dp C, (3.11) provided that β 0 +β + +β + n = α 1 +α, n, and β + +β + n α 1 α 0. The last condition implies that β 0 0 which we shall see that this is not good enough. Lemma 3 (Claim (.10)). Let T c (η, ϕ) = {ξ : ξ η + η, ξ ϕ + ϕ γ ξ }. If β 0 + β + + β + n = α 1 + α and β 0 > n 1, then Dβ 0 Ŝ β + + Ŝβ T c where τ = ξ ϕ + ϕ = ξ η + η. δ( ξ ϕ + ϕ ξ η η ) ( ϕ η ) α 1+α dξ C, (3.1) Proof. Let Φ(ξ) = ξ ϕ + ϕ ξ η η and P (ϕ, η) = {ξ : Φ(ξ) = 0}. Since ξ ϕ + ϕ γ ξ and γ > 16, thus we have ϕ + ξ ϕ 4 ξ and analogously we have η + ξ η 4 ξ. Using triangle inequality, we get 3 5 η ξ η 5 3 η and 3 5 ϕ ξ ϕ 5 3 ϕ, and then ξ min{ η, ξ η, ϕ, ξ ϕ }. 3 On the plane P we have ξ ϕ + ϕ = ξ η + η which implies that ξ ϕ ϕ ξ η η. Set (ξ η) (η) = ξ η η cos θ, through some calculations we can show that cos θ cos θ 0 = 5 3 > = cos π 4. 9

Figure 3: Hence the angle between η and ξ η is restricted on 0 θ θ 0 π 4. Without loss of generality, we assume ϕ > η. Follow the idea used in [FK], we decompose S n = Ω j, where Ω j are disjoint and the angle between any two unit vectors lie N in j=1 the same Ω j is less than θ 0, and N is a finite integer. Denote { Γ j = ξ R n /{0} : ξ ξ Ω j }, χ j = characteristic function of Γ j, f j = χ j f, and g j = χ j g. Thus we have f = pieces, N f j and g = j=1 N g j. Then we can split the integral into finitely many j=1 j,k D β 0 S β + + S β δ(τ ξ η η ) f(ξ η)ĝ(η) dη L (τ 16 ξ ) D β 0 S β + + S β δ(τ ξ η η ) f j (ξ η)ĝ k (η) dη L (τ 16 ξ, θ 0 <π/4). There exists a cone Γ with an aperture θ 0 such that η Γ k Γ and ξ η Γ j Γ. D β 0 S β + + S β δ(τ ξ η η ) f j (ξ η)ĝ k (η) dη L (τ 16 ξ ) = D β 0 S β + δ(φ(ξ)) f j (ξ η)ĝ k (η) fj (ϕ) ĝ k (ξ ϕ) dϕ dη dξ, + S β where η Γ k Γ, ξ η Γ j Γ, ϕ Γ j Γ, and ξ ϕ Γ k Γ. Through elementary argument, we have the following identity, see [H], D β 0 S β + + S β δ(φ(ξ)) ( ϕ η ) α 1+α dξ = 10 D β 0 S β + + S β dµ ( ϕ η ) α 1+α Φ(ξ), (3.13)

Figure 4: where dµ is the surface measure on the surface {ξ : Φ(ξ) = 0}. The facts ξ ϕ Γ and ξ η Γ imply that Φ(ξ) ϕ since Φ(ξ) = (ϕ η) = ϕ + η ϕ η ϕ. Let ξ be the projection of ξ onto the plane P, see figure 5, Figure 5: and P ϕ+η the projection of ϕ + η P ϕ+η ξ = ξ ξ onto the plane P, = ϕ + η ϕ + η ϕ η ϕ η ϕ η ϕ η, ϕ η ϕ η ϕ η ϕ η. Denote the rotation taking ϕ η to ϕ η e 1 by R and the change of coordinates ( ν = R ξ ϕ + η ) = (X 1, X,, X n ). 11

Thus we get ξ ξ and ( ) ( R ξ P ϕ+η = R ξ ϕ + η ) ( R ξ ϕ + η ) ( ) ( ) ϕ η ϕ η R R ϕ η ϕ η = ν ν e 1 e 1 = (X 1, X,, X n ) (X 1, 0,, 0) = (0, X,, X n ), and then dξ = dx dx n = dµ. Hence we obtain δ(φ(ξ))dξ = dµ Φ(ξ) dξ ϕ. Therefore we can now bound the integral (3.13) as follows. ξ β 0 ξ ϕ + ϕ + ξ β + ξ ϕ + ϕ ξ β δ(φ(ξ)) P (η,ϕ), ξ 3 η ( ϕ η ) α 1+α dξ ϕ 4β ++4β α 1 α ξ β dξ 0 ϕ. ξ < ϕ If β 0 0, then we get the bound ξ β 0 dξ ξ < ϕ If β 0 < 0, then we get the bound ξ β 0 dξ ξ ϕ ξ ϕ ξ ϕ ϕ β 0 dξ ϕ β 0+n1. ξ β 0 dξ ϕ β 0+n1, provided that β 0 + n 1 > 0. Finally combining the above results we have (3.13) ϕ 4β ++4β α 1 α +β 0 +n C, provided that β + + β + β 0 + n = α 1 + α and β 0 + n 1 > 0. This completes the proof. Acknowledgement: The author was partially supported by NSC and NCTS(South) in Taiwan. The author also wants to express his gratitude for the hospitality of the Department of Mathematics, Kyoto University, and RIMS during the visit from July 03 010 to July 10 010. References [Be] M. Beals, Self-Spreading and strength of Singularities for solutions to semilinear wave equations, Ann. of Math. 118 (1983), 187-14. [BGT] N. Burq, P. Gerard and N. Tzvetkov, Bilinear eigenfunction estimates and the nonlinear Schrödinger equation on surfaces, Invent. Math. 159 (005), 187-3. 1

[Bo] J. Bourgain, Refinements of Strichartz inequality and applications to D-NLS with critical nonlinearity, Int. Math. Res. Not. 5 (1998) 53-83. [C] T. Cazenave, Semilinear Schrödinger equations, Courant Lecture Notes in Mathematics, 10. New York University, Courant Institute of Mathematical Sciences, New York, 003. [CCL] M. Chae, Y. Cho, and S. Lee, Mixed Norm Estimates of Schrödinger Waves and Their Applications, (010). [CDKS] J. Colliander, J. Delort, C. Kenig, and G. Stafflani, Bilinear Estimates and Applications to D NLS, Transactions AMS 353 (001), 3307-335. [CGT] J. Colliander, M. Grillakis and N. Tzirakis, Improved interaction Morawetz inequalities for the cubic nonlinear Schrödinger equation in d, Int. Math. Res. Not. 3 (007). [CKSTT1] J. Colliander, M. Keel, G. Stafflani, H. Takaoka and T. Tao, Almost conservation laws and global rough solutions to a nonlinear Schrödinger equation, Math. Res. Letters 9 (00), 1-4. [CKSTT], Global existence and scattering for rough solutions of nonlinear Schrödinger equation on R3, Commun. Pure Appl. Math. 57(004), 987-1014. [CKSTT3], Global well-posedness and scattering for the energy-critical nonlinear schrodinger equation in R3, Ann. Math. 167 (008), 767-865. [DFS] P. D Ancona, D. Foschi, and S. Selberg, Null structure and almost optimal local regularity of the Dirac-Klein-Gordon system, Journal of EMS 9 (007), 877-898. [DPST] D. de Silva, N. Pavlovic, G. Staffllani, and N. Tzirakis, Global Well-Posedness for the L-critical nonlinear Schrödinger equation in higher dimensions, Commun. Pure. Appl. Anal. 6(007), 103-1041. [FG] Y. Fang and M. Grillakis, On the global existence of rough solutions of the cubic ocusing Schrödinger equation in R+1, J. Hyperbolic Differ. Equ. 4 (007), 33-57. [FK] D. Foschi and S. Klainerman, Bilinear space-time estimates for homogeneous wave equations, Ann. Sci. E cole Norm. Sup. (4) 33, no. (000), 11-74. [H] [K] L. Hormander, The Analysis of Linear Partial Differential Operators, Grundlehren mathematischen Wissenschaften 56, Springer-Verlag 1983. N. Kishimoto, Low-Regularity Bilinear Estimates for a Quadratic Nonlinear Schrödinger Equation, JDE 47 (009), 1397-1439. [KT] M. Keel and T. Tao, Endpoint Strichartz estimates, Amer. J. Math. 10 (1998), 955-980. [KM] S. Klainerman and M. Machedon, Remark on Strichartz-type inequalities. With appendices by Jean Bourgain and Daniel Tataru, Internat. Math. Res. Notices 5 (1996), 01-0. [KV] S. Keraani and A. Vargas, A smoothing property for the L-critical NLS equations and an application to blowup theory,. to appear in Ann. Inst. H. Poincaré Anal. Non Linéaire, 6 (009), 745-76. [L] S. Lee, Bilinear Restriction Estimates for Surfaces with Curvatures of DiRerent Signs, Trans. Amer. Math. Soc., 358 (006), 3511-3533. [LRV] S. Lee, K. M. Rogers and A. Vargas, Sharp null form estimates for the wave equation in R3+1, Int. Math. Res. Not. IMRN 008. 13

[LV] S. Lee and A. Vargas, Sharp null form estimates for the wave equation, Amer. J. Math. 130 (008), no. 5, 179-136. [OT] T.Ozawa and Y. Tsutsumi, Space-time estimates for null gauge forms and nonlinear Schrödinger equations, Differential and Integral Equations 11 (1998), 01-. [S] R. Strichartz, Restrictions of Fourier transform to quadratic surfaces and decay of solutions of Wave Equations, Duke Math. J. 44 (1977), 705-714. [St] E. M. Stein, Harmonic Analysis: Real-Variable Methods, Orthogonality, and Oscillatory Integrals,. Princeton Math. Ser., vol. 43, Monographs in Harmonic Analysis, III, Princeton University Press, Princeton, NJ, 1993. [T1] T. Tao, A Sharp bilinear restriction estimate for paraboloids, Geom. Funct. Anal. 13 (003), 1359-1384. [T], Nonlinear dispersive equations, Local and global analysis, CBMS 106, eds: AMS, 006. [TVZ] T. Tao, M. Visan and X. Zhang, The Schrödinger equations with combined power-type nonlinearities, Commun. PDE 3 (007), 181-1343. Department of Mathematics, National Cheng Kung University, Tainan, Taiwan and National Center for Theoretical Science email: fang@math.ncku.edu.tw 14